Let $A$ be a $4 \times 4$ matrix over $C$ such that $rank (A)=2$ and $A^3=A^2 \neq 0.$ Suppose that $A$ is not diagonalizable. Then prove that there exist a vector $v$ such that $Av$ is not equal to zero, but $A^2v=0$ and the characteristic polynomial of $A$ is $x^4-x^3$.
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4Some related posts: http://math.stackexchange.com/questions/1086438/true-or-false-statement-about-a-non-diagonal-matrix http://math.stackexchange.com/questions/1295229/show-that-there-exists-a-vector-v-such-that-av-neq-0-but-a2v-0 – Martin Sleziak Jun 14 '16 at 13:15